Which expression is equivalent to 15x + 3?

Did I do the first question right? can anyone please help me with the 2nd question? Would really appreciate the Help

IgorC [24]

Answer:

8. 1483.33

9. 21 months

Step-by-step explanation:

8. Morgan's income after taxes is 55000/12 = 4583.33 per month. The amount available after expenses is 4583.33 -3100.00 = 1483.33 per month.

Morgan is able to put $1483.33 per month into savings.

__

9. If Morgan is able to save $1483.33 per month, it will take her ...

$30,000/$1483.33 ≈ 20.2

months to save $30,000. After 20 months, she won't have quite enough, so it will take her one more month to save the desired amount.

It will take Morgan about 21 months to save $30,000.

_

If you like, you can write an equation for "m", the number of months it will take Morgan to save 30,000:

1483.33×m = 30,000

m = 30,000/1483.33 ≈ 20.2 . . . . . . divide by the coefficient of m

7 0

2 years ago

Let f(x) = x - 3 and g(x) = x + 11 find f(x) times g(x)<br><br> What is the answer?

notsponge [240]

f(x) *g(x) = (x-3)(x+11) = x^2 +11x -3x -33 = x^2 +8x -33

Sorry if its wrong :(

6 0

2 years ago

In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A

vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

$\mu_s=np=500*0.90=450$

The standard deviation will be:

$\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7$

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

$z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932$

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

$\mu=np=300*0.932=279.6$

The standard deviation will be:

$\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36$

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

$P(270\leq x\leq280)=P(269.5$